5
$\begingroup$

In "Quantum Field Theory" by Mark Srednicki, Chapter 9 page 67, after he proves that $\langle 0|\phi(x)|0 \rangle$ vanishes (meaning sum of all connected diagrams with a single source is zero), he makes the following claim:

Consider now that same infinite set of diagrams, but replace the single source in each of them with some other subdiagram. Here is the point: no matter what this replacement subdiagram is, the sum of all these diagrams is still zero. Therefore we need not bother to compute any of them! The rule is this: ignore any diagram that, when a single line is cut, falls into two parts, one of which has no sources. All of these diagrams (known as tadpoles) are canceled by the $Y$ counterterm, no matter what subdiagram they are attached to.

The most important question that I am wondering is how he arrived at this conclusion from his proof that the $Y$ counterterm can be used to make $\langle0|\phi(x)|0\rangle$ zero.

Also, what does he mean by "subdiagram" here? A part of the diagram formed by cutting one of the diagram that has a source, or a part of any diagram that does not necessarily have a source? Is he replacing each of the different diagrams with single source with identical subdiagram or replacing each source with different subdiagrams? (Since "subdiagram" is singular, I am guessing they are all replaced with identical subdiagrams.)

$\endgroup$
2
$\begingroup$

Comments to the question (v1):

  1. Ref.1 is considering $\varphi^3$ theory $$\tag{1} {\cal L}(J)~=~\frac{1}{2}Z_{\varphi}\partial^{\mu}\varphi\partial_{\mu}\varphi - \frac{1}{2}Z_{m}m^2\varphi^2 - \frac{1}{6}Z_{g}g\varphi^3+(Y+J)\varphi.$$ To read the Feynman diagrams in Ref. 1, note that the source $J(x)$ is drawn as a black bullet $\bullet$, and the counterterm $Y(x)$ is drawn as a cross $\times$.

  2. Technically, $$\tag{2} \langle 0 | \varphi(x) | 0 \rangle~=~ \frac{1}{i} \left. \frac{\delta W(J)}{\delta J(x)}\right|_{J=0} $$ is the sum of all connected Feynman diagrams with a single source $J(x)$, with the source removed/striped.

  3. We have adjusted the $Y$ counterterm in the $\varphi^3$ theory, so that the sum (2) is zero: $$\tag{3} \langle 0 | \varphi(x) | 0 \rangle~=~0.$$

  4. We now consider the same collection of Feynman diagrams, with the single source $J(x)$ replaced by a fixed but arbitrary subdiagram $SD(x)$. The corresponding sum will then again vanish $$\tag{4} \int\!d^4x~ SD(x)\langle 0 | \varphi(x) | 0 \rangle~=~0,$$ due eq. (3) and the distributive law.

References:

  1. M. Srednicki, QFT, Chapter 9. A prepublication draft PDF file is available here.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.